If P = begin{vmatrix} 1 & alpha & 3 1 & 3 & | vedclass

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(B) Given that $P = \operatorname{adj}(A)$ and $\det(A) = 4$.We know that for a $3 	imes 3$ matrix $A$, $|\operatorname{adj}(A)| = |A|^{n-1}$, where

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$11$

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(B) Given that $P = \operatorname{adj}(A)$ and $\det(A) = 4$.We know that for a $3 	imes 3$ matrix $A$, $|\operatorname{adj}(A)| = |A|^{n-1}$, where $n=3$.Therefore, $|P| = |A|^{3-1} = |A|^2 = (4)^2 = 16$.Now, calculate the determinant of $P$:$|P| = \begin{vmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{vmatrix}$$|P| = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$$|P| = 0 - \alpha(-2) + 3(-2)$$|P| = 2\alpha - 6$.Equating the two values of $|P|$:$2\alpha - 6 = 16$$2\alpha = 22$$\alpha = 11$.

If $P = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $\det(A) = 4$, then $\alpha$ is equal to

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